leetcode 448. Find All Numbers Disappeared in an Array 寻找缺失元素+很棒O(n)做法

我就是我 2022-06-04 08:39 173阅读 0赞

Given an array of integers where 1 ≤ a\[i\] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements of \[1, n\] inclusive that do not appear in this array.

Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

Example:

Input:  
\[4,3,2,7,8,2,3,1\]

Output:  
\[5,6\]

我最初的想法是就是使用set对重复元素做过滤，然后遍历查询即可，这个是另一个很不错的想法

这种解法的思路路是，对于每个数字nums\[i\]，如果其对应的nums\[nums\[i\] - 1\]是正数，我们就赋值为其相反数，如果已经是负数了，就不变了，那么最后我们只要把留下的整数对应的位置加入结果res中即可，

建议和[leetcode 442. Find All Duplicates in an Array 重复元素查找+很棒O(n)做法][leetcode 442. Find All Duplicates in an Array _O_n] 一起学习

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <cmath>
    
    using namespace std;
    
    class Solution 
    {
    public:
        vector<int> findDisappearedNumbers(vector<int>& num) 
        {
            for (int i = 0; i < num.size(); i++)
            {
                int index = abs(num[i]) - 1;
                if (num[index] > 0)
                    num[index] = -num[index];
            }
    
            vector<int> res;
            for (int i = 0; i < num.size(); i++)
            {
                if (num[i] > 0)
                    res.push_back(i+1);
            }
            return res;
        }
    };

[leetcode 442. Find All Duplicates in an Array _O_n]: http://blog.csdn.net/jackzhang_123/article/details/78753911